If $\alpha$ is the degree of dissociation of ${\text{Na}}_2{\text{SO}}_4$, the van't Hoff's factor ($i$) used for calculating the molecular mass is:

${\text{Na}}_2{\text{SO}}_4$ dissociates as follows:
$${\text{Na}}_2{\text{SO}}_4\rightleftharpoons 2{\text{Na}}^{+}+{\text{SO}}_4^{2-}$$
Let the initial moles be $1$.Initial: $1$ mole of ${\text{Na}}_2{\text{SO}}_4$, $0$ moles of ions.At equilibrium: $(1-\alpha )$ moles of ${\text{Na}}_2{\text{SO}}_4$, $2\alpha$ moles of ${\text{Na}}^{+}$, and $\alpha$ moles of ${\text{SO}}_4^{2-}$.Total number of moles after dissociation:
$$i=(1-\alpha )+2\alpha +\alpha =1+2\alpha$$
Thus, the van't Hoff factor $i=1+2\alpha$.